Looking to reduce a given voltage to a specific amperage

Status
Not open for further replies.

Boosted67

Junior Member level 1
Joined
Apr 30, 2014
Messages
16
Helped
0
Reputation
0
Reaction score
0
Trophy points
1
Visit site
Activity points
175
I know voltage isn't amperage, but I have either a 12v or 5v source to choose from. This will be feeding a varying solenoid. The position I'd like it in will require 850 milliamps(say the instructions.) Since this isn't a simple voltage reduction I was wondering if you guys had any input. It would be even better if I could vary the amperage, just incase I need the adjustment to get the solenoid right where I want it. Thanx for any direction in advance
 

There is no such word as 'Amperage' but I think you want to limit the current to 850mA.

The voltage and current are related by the resistance of the solenoid. Although a higher voltage will result in a higher current, to know how to limit it to 850mA you first need to know the solenoids resistance. There are various methods to limit the current but it may be a simple as adding a fixed resistor in series with the solenoid. The calculation of the resistor value is:

(supply voltage / 0.85) - solenoid resistance. The result is in Ohms.

Example: solenoid resistance is 10 Ohms, voltage is 12V. The resistor needed is (12/0.85)-10 = 4.1 Ohms

Brian.
 
No such word as amperage you say. Shows you my level of "expertise" eh? Lol

I went back and looked, and it actually looks like 50mA is what it needs. Solenoid is going to be 5.5ohms. And a 5v source is going to make my resistor smaller, so,....

(5/.05)-5.5=94.5 ohm resistor. I can buy a pot in that range and be able to adjust to get her exactly where I need? That simple?

Can I feed power directly to the solenoid, and put the resistor on the negative side afterwards, and achieve the same result? Limiting the ground would also limit the power thru the solenoid right?
 

"Amperage" is used in a colloquial sense through-out the electrical / electronics industry.... for decades...

You need a circuit that will supply a constant (ideally adjustable) current to your solenoid, preferably a switcher so that little power is wasted from your sources, there are numerous ways of doing this, even with standard switching ic's from national and the like... or the 33063 series
 
Boosted67, yes in theory that is correct and it doesn't matter which side of the solenoid you connect the resistor.

Anna Conda is also correct although the use of methods other than a simple resistor depends on the circumstances. A resistor will always work but as a product of dropping the current it also produces heat. The power lost in the 94.5 Ohm resistor will be (I^2 * R) = .05*.05 * 94.5 = 0.24W which is quite small and even a physically small resistor would suffice. Be careful if you use a potentiometer/variable resistor though, if you do, add a fixed resistor in series with it, for example 75 Ohm fixed and 50 Ohm variable. It will still give you adjustment around the value you want but the heat will be shared between the fixed and variable resistors. If you use a pot alone, the heat is only dissipated in the part of the track carrying the current which might be short enough for it to overheat.

For larger currents a resistor will still work but the circuit can be made more efficient by using switching methods to limit it.

Please check the solenoid ratings again, I'm not saying they are wrong but it would be more usual for one with 5.5 Ohm resistance to operate at a higher current than 50mA.

Brian.
 
Thanks a bunch guys, really does help. It will be low use item. I think the low cost and rock solid reliablity of a resistor should work out awesome. Two resistors is an awesome idea. Should have thought of that for the last project when I was doing higher current and had to get a big resistor and epoxy to a heat sink lol

You're right Brian, the numbers seem odd for the solenoid, but it has an inverse operation.

Thanx again for the help guys
 

A transistor is a simple and easy way to regulate current.

Supply ranges between 5 and 12V.
The 5 ohm load draws the same milli-Amps (within a few percent).



It requires a stable bias. The darlington configuration helps in this purpose.

Several watts of heat are wasted. However if the solenoid is used only for short periods, then heat may not be serious.
 

Hmmm @Brad, as the xtors heat up their gain will change, there are a couple of other constant current ckts that are just as simple and give much more stable current over temp and at cold start....
 

Status
Not open for further replies.
Cookies are required to use this site. You must accept them to continue using the site. Learn more…